Problem: There is a railroad in Takahashi Kingdom. The railroad consists of N sections, numbered 1, 2, ..., N, and N+1 stations, numbered 0, 1, ..., N. Section i directly connects the stations i-1 and i. A train takes exactly A_i minutes to run through section i, regardless of direction. Each of the N sections is either single-tracked over the whole length, or double-tracked over the whole length. If B_i = 1, section i is single-tracked; if B_i = 2, section i is double-tracked. Two trains running in opposite directions can cross each other on a double-tracked section, but not on a single-tracked section. Trains can also cross each other at a station.

Snuke is creating the timetable for this railroad. In this timetable, the trains on the railroad run every K minutes, as shown in the following figure. Here, bold lines represent the positions of trains running on the railroad. (See Sample 1 for clarification.)

<image>

When creating such a timetable, find the minimum sum of the amount of time required for a train to depart station 0 and reach station N, and the amount of time required for a train to depart station N and reach station 0. It can be proved that, if there exists a timetable satisfying the conditions in this problem, this minimum sum is always an integer.

Formally, the times at which trains arrive and depart must satisfy the following:

* Each train either departs station 0 and is bound for station N, or departs station N and is bound for station 0.
* Each train takes exactly A_i minutes to run through section i. For example, if a train bound for station N departs station i-1 at time t, the train arrives at station i exactly at time t+A_i.
* Assume that a train bound for station N arrives at a station at time s, and departs the station at time t. Then, the next train bound for station N arrives at the station at time s+K, and departs the station at time t+K. Additionally, the previous train bound for station N arrives at the station at time s-K, and departs the station at time t-K. This must also be true for trains bound for station 0.
* Trains running in opposite directions must not be running on the same single-tracked section (except the stations at both ends) at the same time.

Constraints

* 1 \leq N \leq 100000
* 1 \leq K \leq 10^9
* 1 \leq A_i \leq 10^9
* A_i is an integer.
* B_i is either 1 or 2.

Input

The input is given from Standard Input in the following format:


N K
A_1 B_1
A_2 B_2
:
A_N B_N


Output

Print an integer representing the minimum sum of the amount of time required for a train to depart station 0 and reach station N, and the amount of time required for a train to depart station N and reach station 0. If it is impossible to create a timetable satisfying the conditions, print -1 instead.

Examples

Input

3 10
4 1
3 1
4 1


Output

26


Input

1 10
10 1


Output

-1


Input

6 4
1 1
1 1
1 1
1 1
1 1
1 1


Output

12


Input

20 987654321
129662684 2
162021979 1
458437539 1
319670097 2
202863355 1
112218745 1
348732033 1
323036578 1
382398703 1
55854389 1
283445191 1
151300613 1
693338042 2
191178308 2
386707193 1
204580036 1
335134457 1
122253639 1
824646518 2
902554792 2


Output

14829091348

Solution (in CPP):

Solution:
#include<stdio.h>
#include<math.h>
#include<algorithm>
#include<queue>
#include<deque>
#include<string>
#include<string.h>
#include<vector>
#include<set>
#include<map>
#include<stdlib.h>
using namespace std;
const long long mod=1000000007;
const long long inf=mod*mod;
const long long d2=500000004;
int c[110000];
int d[110000];
long long segtree[262144];
long long query(int a){
	long long ret=0;
	a+=131072;
	while(a){
		ret=max(ret,segtree[a]);
		a/=2;
	}
	return ret;
}
void update(int a,int b,int c,int d,int e,long long f){
	if(d<a||b<c)return;
	if(c<=a&&b<=d){
		segtree[e]=max(segtree[e],f);
		return;
	}
	update(a,(a+b)/2,c,d,e*2,f);
	update((a+b)/2+1,b,c,d,e*2+1,f);
}
long long dp[110000];
long long x[110000];
long long z[110000];
 
long long ans[2][110000];
int main(){
	int a,b;scanf("%d%d",&a,&b);
	for(int i=0;i<a;i++){
		scanf("%d%d",c+i,d+i);
		c[i]*=2;
		if(d[i]==1&&c[i]>b){
			printf("-1\n");return 0;
		}
	}
	for(int t=0;t<2;t++){
		for(int i=0;i<a;i++){
			x[i+1]=x[i]+c[i];
		}
		for(int i=0;i<=a;i++){
			z[i]=x[i]%b;
		}
		std::sort(z,z+a+1);
		for(int i=0;i<262144;i++)segtree[i]=0;
		for(int i=0;i<110000;i++)ans[t][i]=inf;
		ans[t][0]=0;
		for(int i=0;i<a;i++){
			int at=lower_bound(z,z+a+1,x[i+1]%b)-z;
			int la=query(at);
			if(la==0){
				ans[t][i+1]=x[i+1];
			}else{
				ans[t][i+1]=ans[t][la]+(long long)((x[i+1]-x[la]+b-1)/b)*b;
			}
 
			if(d[i]==1){
				long long L=(x[i]+1)%b;
				long long R=x[i+1]%b;
				int l=lower_bound(z,z+a+1,L)-z;
				int r=lower_bound(z,z+a+1,R)-z;
				if(L>R){
					update(0,131071,0,r,1,i+1);
					update(0,131071,l,131071,1,i+1);
				}else{
					update(0,131071,l,r,1,i+1);
				}
			}
		//	printf("%d: %lld\n",i+1,ans[t][i+1]);
		}
 
		reverse(c,c+a);
		reverse(d,d+a);
	}
	long long ret=inf;
	for(int i=0;i<=a;i++){
		ret=min(ret,ans[0][i]+ans[1][a-i]);
	}
 
	printf("%lld\n",ret);
}